6.6.1-eg-1

# 6.6.1-eg-1 - k According to Hooke’s Law the force F is modeled using F x = kx Since we wish to stretch the spring z = 0 80 m beyond its natural

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Example It took 45 J of work to stretch a spring from its natural length of 20 cm to a length of 1 m. Find the spring’s force constant. (a) Convert length measurements to meters. We will measure work in Joules (J), force in Newtons (N), and displacement in meters (m) in order to employ the relation J = N-m. Since 1 m = 100 cm, it follows that 20 cm = 20 cm * 1 m 100 cm = 0 . 20 m . (b) Use the formula for work and Hooke’s Law to ﬁnd
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Unformatted text preview: k . According to Hooke’s Law, the force F is modeled using F ( x ) = kx . Since we wish to stretch the spring z = 0 . 80 m beyond its natural length, we have that the work W satisﬁes W = Z z F ( x ) dx = Z . 80 kx dx = 45 . Solving for k in 45 = kx 2 2 ± ± ± ± ± . 80 = k (0 . 80) 2 2 = k (0 . 32) , it follows that k = 45 . 32 = 140 . 625 ....
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## This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.

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